Now position vector of C with the parameters of c,k in the Plane equation.
My thinking: the length of AC is twice as long as the length AM. So i came to the conclusion my parameters should be

c = 0 since i dont need to go in the direction of point B
and d = 2 since its twice as long. Really not sure how to solve that more delicate

Question 3: Find the vector v perpendicular to the Plane ABM, such that it is pointing downwards

I think i need something like a Normalvector found by a crossproduct. But im confused since i thought i needed only 2 vectors

for that....

Question 4: Find the cosine of the angle between the lines AB and AC at the point A

cos(phi) = [ (v1 * v2) / (|v1| x |v2|) ]

v1 = AB = [3;-2;3] +c [4;0;1]
v2 = AC = A2M

not sure about that last one

to #2): You don't need the equation of the plane to find the coordinates of point C.
You know that the mean of the position vectors of A and C is the position vector of M. Let $\displaystyle C(x_C, y_C, z_C)$ then

$\displaystyle \frac12([3, -2, 3] + [x_C, y_C, z_C]) = [6,3,7]$

You'll get C(9, 8, 11).

to #3): The orientation of the plane in 3-D is produced by the direction vectors of the plane. Therefore